Sections 8.1, 8.2, 8.3, 8.4 and 8.5
The Moon's craters are remnants of meteorite collisions. Suppose a fairly large asteroid that has a mass of 5.00×10^12kg5.00×1012kg (about a kilometer across) strikes the Moon at a speed of 15.0 km/s. At what speed does the Moon recoil after the perfectly inelastic collision (the mass of the Moon is 7.36×1022kg7.36×1022kg)?
(5x10^12 x 15x10^3)/(5x10^12+7.36x10^22)=1.02×10−6 m/s
With a mass of 111.1kg, m1 is moving to the right with a velocity of 3.000m/s.m2 is moving at 5.52m/s to the LEFT and has a mass of 222.2kg.After the elastic collision, mass1 bounces off with a velocity of 8.000m/s to the LEFT.Calculate the magnitude of the velocity of m2 after the collision, in m/s.
0.0200
How much kinetic energy is lost in the collision? Such an event may have been observed by medieval English monks who reported observing a red glow and subsequent haze about the Moon.
1/2(5x10^12 + 7.36 x 10^22) x (1.019x10^-6)^2 - 1/2(5x10^12) x (15 x 10^3)^2 = 5.63×10^20 J
Select all of the following real life situations that approximate an elastic collision: 1car runs into a brick wall 2cue ball strikes another billiard ball 3two oxygen molecules bump into each other, form O2, and travel in the direction the faster moving molecule was headed
2
A 7.77 kg pumpkin falls from a height of 12.3 m. It hits the ground and comes to an abrupt stop in about 444 ms. What is the average force imparted by the ground on the pumpkin? [HINT: Use the conservation of energy to find the velocity of the pumpkin at the moment just before it hits the ground.]
272 N v=sqrt 2gh = sqrt 2 x 9.8 x 12.3 v=15.53 t=444ms = 0.444 sec a=v/a=35 m/s f=ma=(7.77 x 35) = 271.72N
Which form of Newton's second law would you use to solve a problem in which mass varies?
Fnet = deltaP/delta t The net external force will be given by the rate of change of momentum.
What is the equation for Newton's second law in terms of momentum?
Fnet=ΔpΔt According to Newton's second law, net external force is the rate at which momentum changes
Tennis racquets have "sweet spots." If the ball hits a sweet spot then the player's arm is not jarred as much as it would be otherwise. How does the "sweet spot" change the impact time of the ball?
It increases impact time
State the law of conservation of momentum.
Momentum is conserved for an isolated system with any number of objects in it.
Two steel balls collide and bounce apart in an isolated system. If the final kinetic energy is less than the initial kinetic energy, which of the following is true?
Momentum is conserved in this inelastic collision
What is the difference between momentum and impulse?
Momentum is the product of mass and velocity. Impulse is the change in momentum.
Are perfectly elastic collisions possible?
Perfectly elastic collisions are possible only with subatomic particles.
Two identical billiard balls strike a rigid wall with the same speed, and are reflected without any change of speed. The first ball strikes perpendicular to the wall. The second ball strikes the wall at an angle of 30º from the perpendicular, and bounces off at an angle of 30º from perpendicular to the wall. (a) Determine the direction of the force on the wall due to each ball. (b) Calculate the ratio of the magnitudes of impulses on the two balls by the wall.
These changes mean the change in momentum for both balls is in the −𝑥−x direction, so the force of the wall on each ball is along the −𝑥−x direction. Impulse is the change in momentum vector. Therefore the x comp is -2mu and the 𝑦 is 0 2mu/2mucos30 = 2/sqrt 3 = 1.155
collision: Cart 1 has a mass of 0.350kg and an initial velocity of 2.00m/s. Cart 2 has a mass of 0.500kg and an initial velocity of −0.500m/s. After the collision, cart 1 recoils with a velocity of −4.00m/s. What is the final velocity of cart 2?
Total momentum before the collision is 0.45kg⋅m/s, so to conserve momentum the final velocity of cart 2 is 3.7m/s.
How can you tell that an object is in motion?
When there is a change in the background and position of the moving object with respect to the frame of reference.
What is a perfect inelastic collision?
a A perfect inelastic collision is one in which objects stick together after impact, and their internal energy is not conserved.
In the equation 𝑝1+𝑝2=𝑝′1+𝑝′2p1+p2=p1′+p2′ for the collision of two objects, what is the assumption made regarding the friction acting on the objects?
a Friction is zero.
A Saturn V's mass at liftoff was 2.80×10^6kg, its fuel burn rate was 1.40×10^4 kg/s and the exhaust velocity was 2.40x10^3 m/s. Calculate its initial acceleration.
a=ve/m * (deltam/deltat) - g =((2.4x10^3 m/s)/(2.8x10^6))(1.40 x 10^4 kg/s)-9.8m/s^2 =2.2 m/s^2
Through what means is kinetic energy lost in collisions between everyday objects?
due to friction experienced by the objects
Calculate the momentum of a 110-kg football player running at 8.00 m/s. (b) Compare the player's momentum with the momentum of a hard-thrown 0.410-kg football that has a speed of 25.0 m/s.
p=mv player=(110 kg)(8.00 m/s)=880 kg·m/s ball=(0.410 kg)(25.0 m/s)=10.3 kg·m/s player/ball=880/10.3=85.9
A 50 g hockey puck is slapped with an initial velocity of 50 m/s at an angle of 60° from the x-axis. What is the x-component of the momentum, in kg·m/s?
p=mvcos theta = 0.050 kg x (50m/s) x cos60 = 1.25 kg·m/s
How can you state the law of conservation of momentum in equation form?
ptot=p'tot
Suppose the following experiment is performed. A 0.250-kg object (m1) is slid on a frictionless surface into a dark room, where it strikes an initially stationary object with mass of 0.400 kg (m2) 45.0º with its incoming direction. The speed of the 0.250-kg object is originally 2.00 m/s and is 1.50 m/s after the collision. Calculate the magnitude and direction of the velocity (v′2 and θ2) of the 0.400-kg object after the collision.
tan θ2 = ((1.50m/s)(0.7071))/((1.50m/s)(0.7071)-2m/s) =-1/129 θ2=tan^-1(−1.129)=311.5º≈312º. v'2=-(.250kg/.4kg)(1.5m/s)(0.7071/-0.7485)
Momentum in the two-ball system is conserved when
the two balls collide with each other but not when they collide with the walls.
To calculate the speed of the pebble dropped from the cliff as it hits the ground requires you to only know the height of the cliff and acceleration due to gravity.
true
True or False: Devices with efficiencies of less than one do not conserve their mechanical energy.
true
Cart 1 (denoted m1 carries a spring which is initially compressed. During the collision, the spring releases its potential energy and converts it to internal kinetic energy. The mass of cart 1 and the spring is 0.350 kg, and the cart and the spring together have an initial velocity of 2.00 m/s has a mass of 0.500 kg and an initial velocity of −0.500 m/s. After collision, cart 1 recoil w velocity −4.00 m/s. cart 2 vf? amount of energy released by spring?
v'2=(m1v1+m2v2-m1v'1)/m^2 =(((0.350kg)(2m/s)+(0.50kg)(-0.5m/s))/0.5kg)-(((0.350kg))(-4m/s)/0.5kg)) =3.70m/s ke int = 1/2m1v1^2+1/2m2v2^2 =1/2(0.350kg)(2.0m/s)^2+1/2(0.5kg)(-0.5m/s)^2 =0.763J ke' int = 1/2m1v'1^2+1/2m2v'2^2 =1/2(0.35kg)(-4m/s)^2+1/2(0.5kg)(3.7m/s)^2 =6.22J ke'int - keint = 6.22J - 0.763 J = 5.46 J
Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible.
v'=(0.150 kg/0.150 kg + 70 kg)(35 m/s)=7.48 x 10^-2 m/s ke int = 1/2mv^2 = 1/2(.150 kg)(5 m/s)^2 = 91.9J KE ' int = 1/2(m+M)v^2=1/2(70.15kg)(7.48x10^-2m/s)^2 = 0.196 J ke ' int - ke int = 0.196J-91.9J=-91.7J
How can you express impulse in terms of mass and velocity when neither of those are constant?
Δ𝑝=Δ(𝑚𝑣)
During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women's match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams' racquet, assuming that the ball's speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?
Δ𝑝=𝑚(𝑣f-𝑣i)(0.057 kg)(58 m/s-0 m/s) =3.306 kg·m/s ≈3.3 kg·m/s
Suppose a bicyclist rides at a constant velocity of 4.4 m/s up a 10° slope. The total mass of bicycle and rider is 85 kg. Neglecting air friction, what is his power output?
𝑃=𝑚𝑔𝑣sin𝜃=(85kg)(9.8m/s2)(4.4m/s)sin10∘=0.64kW
Calculate the velocities of two objects following an elastic collision, given that 𝑚1=0.500 kg, 𝑚2=3.50 kg, 𝑣1=4.00 m/s, and 𝑣2=0.m
𝑣′2= 𝑚1/𝑚2(𝑣1−𝑣′1) =0.500 kg/ 3.50 kg[4.00−(-3.00)]m/s v′2=1 m/s
